1033 To Fill or Not to Fill

With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are asked to carefully design the cheapest route to go.

Input Specification:

Each input file contains one test case. For each case, the first line contains 4 positive numbers: C​max​​ (≤ 100), the maximum capacity of the tank; D (≤30000), the distance between Hangzhou and the destination city; D​avg​​ (≤20), the average distance per unit gas that the car can run; and N (≤ 500), the total number of gas stations. Then N lines follow, each contains a pair of non-negative numbers: P​i​​, the unit gas price, and D​i​​ (≤), the distance between this station and Hangzhou, for ,. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the cheapest price in a line, accurate up to 2 decimal places. It is assumed that the tank is empty at the beginning. If it is impossible to reach the destination, print The maximum travel distance = X where X is the maximum possible distance the car can run, accurate up to 2 decimal places.

Sample Input 1:

50 1300 12 8
6.00 1250
7.00 600
7.00 150
7.10 0
7.20 200
7.50 400
7.30 1000
6.85 300

 

Sample Output 1:

749.17

 

Sample Input 2:

50 1300 12 2
7.10 0
7.00 600

 

Sample Output 2:

The maximum travel distance = 1200.00

题意：

　　汽车刚开始的时候没有油，图中会经过一些加油站，每个加油站的油价不同，问选择那些加油站加油，会使全程的油价最低。

思路：

　　这道题用到的是贪心算法，在选择加油的过程中会出现两种情况：

　　1. 从当前加油站向前行驶，在途中会遇到比当前加油站油价更低的加油站，这时我们只需要在当前加油站加的油能够跑到下一个加油站即可。

　　2.从当前加油站向前行驶，当油箱中的廉价油都消耗完了之后，仍然没有找到比当前加油站油价更低的加油站，这时我们要将油箱加满，然后寻找途中相对廉价的加油站加油。

Code：

#include <bits/stdc++.h>

using namespace std;

const int inf = 0x7fffffff;

struct Station {
    double price;
    double distance;
} sta[505];

bool cmp(Station a, Station b) { return a.distance < b.distance; }

int main() {
    int n;
    double c, dis, avg;
    cin >> c >> dis >> avg >> n;
    double price, distance;
    for (int i = 1; i <= n; ++i) {
        cin >> price >> distance;
        sta[i].price = price;
        sta[i].distance = distance;
    }
    sta[0] = {0.0, dis};
    sort(sta, sta + n + 1, cmp);
    double nowPrice = 0.0, totalprice = 0.0, leftdis = 0.0, nowdis = 0.0,
           maxdis = 0.0;
    // 如果第一个加油站不在起点的话，则直接结束
    if (sta[0].distance != 0) {
        cout << "The maximum travel distance = 0.00" << endl;
        return 0;
    } else {
        nowPrice = sta[0].price;
    }
    while (nowdis < dis) {
        maxdis = nowdis + c * avg;
        // 用来标记在所能达到的历程内是否有比当前加油站更便宜的汽油
        int flag = 0;
        // 在所能行驶的里程内，最便宜油价的加油站的距离和油价
        // 如果 minPrice = inf 则表示该加油站为最后一个加油站
        double minPirceDis = 0.0, minPrice = inf;
        // 从第二个加油站开始遍历，贪心寻找最便宜的加油站，使整个旅程的价格最低
        for (int i = 1; i <= n && sta[i].distance <= maxdis; ++i) {
            if (sta[i].distance <= nowdis) continue;
            // 在所能达到的历程中找到比当前油价更便宜的加油站
            if (nowPrice > sta[i].price) {
                flag = 1;
                // 在当前加油站加油，使tank中的油刚好能够到达比当前油价便宜的下一个加油站
                totalprice +=
                    (sta[i].distance - nowdis - leftdis) / avg * nowPrice;
                nowdis = sta[i].distance;
                nowPrice = sta[i].price;
                leftdis = 0.0;
                break;
            }
            // 寻找在所能达到的历程中，油价最便宜的加油站
            if (sta[i].price < minPrice) {
                minPrice = sta[i].price;
                minPirceDis = sta[i].distance;
            }
        }
        // 在当前所能达到的历程中，不存在比当前更便宜的加油站时
        if (flag == 0 && minPrice != inf) {
            // 因为不存在比当前油价更能便宜的加油站，所以应该在当前油站加满油箱，到达历程中油价最便宜的那个油站
            totalprice += nowPrice * (c - leftdis / avg);
            // 更新油箱中剩余油量所能行驶的距离
            leftdis = c * avg - (minPirceDis - nowdis);
            nowPrice = minPrice;
            nowdis = minPirceDis;
        }
        // 当前加油站为最后一个加油站,且没有到达目的地
        if (flag == 0 && minPrice == inf) {
            nowdis += c * avg;
            cout << "The maximum travel distance = " << fixed << setprecision(2)
                 << nowdis << endl;
            return 0;
        }
    }
    cout << fixed << setprecision(2) << totalprice << endl;
    return 0;
}

参考：

　　https://www.liuchuo.net/archives/2461